1
MHT CET 2023 9th May Morning Shift
+1
-0

The excess of pressure in a first soap bubble is three times that of other soap bubble. Then the ratio of the volume of first bubble to other is

A
$$1: 3$$
B
$$27: 1$$
C
$$1: 9$$
D
$$1: 27$$
2
MHT CET 2023 9th May Morning Shift
+1
-0

The radii of two soap bubbles are $$r_1$$ and $$r_2$$. In isothermal condition they combine with each other to form a single bubble. The radius of resultant bubble is

A
$$\mathrm{R}=\frac{\mathrm{r}_1+\mathrm{r}_2}{2}$$
B
$$\mathrm{R}=\mathrm{r}_1\left(\mathrm{r}_1 \mathrm{r}_2+\mathrm{r}_2\right)$$
C
$$\mathrm{R}=\sqrt{\mathrm{r}_1^2+\mathrm{r}_2^2}$$
D
$$\mathrm{R}=\mathrm{r}_1+\mathrm{r}_2$$
3
MHT CET 2022 11th August Evening Shift
+1
-0

The work done in blowing a soap bubble of radius $$\mathrm{R}$$ is '$$\mathrm{W}_1$$' at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius $$2 \mathrm{R}$$ is blown and the work done is '$$\mathrm{W}_2$$'. Then

A
$$\mathrm{W}_2=\mathrm{W}_1$$
B
$$\mathrm{W}_2=4 \mathrm{~W}_1$$
C
$$\mathrm{W}_2<4 \mathrm{~W}_1$$
D
$$\mathrm{W}_2=0$$
4
MHT CET 2022 11th August Evening Shift
+1
-0

A steel coin of thickness '$$\mathrm{d}$$' and density '$$\rho$$' is floating on water of surface tension '$$T$$'. The radius of the coin $$(R)$$ is [$$\mathrm{g}=$$ acceleration due to gravity]

A
$$\frac{\mathrm{T}}{\rho \mathrm{gd}}$$
B
$$\frac{4 \mathrm{~T}}{3 \rho g \mathrm{~d}}$$
C
$$\frac{3 \mathrm{~T}}{4 \rho \mathrm{gd}}$$
D
$$\frac{2 \mathrm{~T}}{\rho \mathrm{gd}}$$
EXAM MAP
Medical
NEET